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Math Help - Work and Fluid Force

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    Senior Member vaironxxrd's Avatar
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    Work and Fluid Force

    An upright cylindrical tank is 10 feet in diameter and 10 feet high. If water in the tank is 6 feet deep. How much work is done in pumping all the water over the top edge of the tank?
    I drew a picture in my notebook and tried to simplify the process, although I'm pretty sure as usual I'm doing most/all steps incorrectly.

    The radius must be 5ft.

    While this exercise doesn't list the weight of the water the whole chapter uses \delta = 62.4 in pounds per cubic foot.

    I'm reading examples which use cones instead of cylinders and literary guessing the following equations...

    F = weight of water * distance

    Weight of water = 62.4, distance = 6-y

    F = \delta\pi(6-y)

    W = F * D

    I'm not sure if the integral should go from 0 to 10 or 0 to 6

    \delta\pi(6-y) * (6-y)dy

    I know this is a big mess, and I'm doing this horribly wrong but I just wanted to give my attempts before asking for help.

    P.S I don't know what's wrong with these equations
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    Re: Work and Fluid Force

    Quote Originally Posted by vaironxxrd View Post
    An upright cylindrical tank is 10 feet in diameter and 10 feet high. If water in the tank is 6 feet deep. How much work is done in pumping all the water over the top edge of the tank?
    I drew a picture in my notebook and tried to simplify the process, although I'm pretty sure as usual I'm doing most/all steps incorrectly.

    The radius must be 5ft.

    While this exercise doesn't list the weight of the water the whole chapter uses \delta = 62.4 in pounds per cubic foot.

    I'm reading examples which use cones instead of cylinders and literary guessing the following equations...

    F = weight of water * distance

    Weight of water = 62.4, distance = 6-y

    F = \delta\pi(6-y)
    It would help if you would be more specific about what these mean. "y" is the height in the tank, measured from the bottom, isn't it? The water has to be lifted to the top of the tank so from height y to 10 feet. The height a "layer of water" has to be lifted is 10- y, not 6- y.

    W = F * D

    I'm not sure if the integral should go from 0 to 10 or 0 to 6

    \delta\pi(6-y) * (6-y)dy

    I know this is a big mess, and I'm doing this horribly wrong but I just wanted to give my attempts before asking for help.

    P.S I don't know what's wrong with these equations
    The 'tex' tags didn't work for me either so I took them out.

    A "layer of water" of height dy has volume Area*height= \pi(5)^2*dx= 25\pi dx and so weight 25\pi \delta dy. It has to be lifted distance 10- y so the work done lifting one "layer" is 25\pi (10- y)dy. Since the water to be lifted goes from y= 0 to y= 6, integrate between 0 and 6.
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    Re: Work and Fluid Force

    It's the work necessay to lift the water from it's cg to the top of tank: W=wh

    Or integrate over the work for each slice using the same formula.
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    Re: Work and Fluid Force

    Quote Originally Posted by HallsofIvy View Post
    It would help if you would be more specific about what these mean. "y" is the height in the tank, measured from the bottom, isn't it? The water has to be lifted to the top of the tank so from height y to 10 feet. The height a "layer of water" has to be lifted is 10- y, not 6- y.


    The 'tex' tags didn't work for me either so I took them out.

    A "layer of water" of height dy has volume Area*height= \pi(5)^2*dx= 25\pi dx and so weight 25\pi \delta dy. It has to be lifted distance 10- y so the work done lifting one "layer" is 25\pi (10- y)dy. Since the water to be lifted goes from y= 0 to y= 6, integrate between 0 and 6.
    As the person who supplied that I feel ashamed to say I don't know what I'm doing at all. I read an example using a cone rather than a cylinder. All of the other examples in the book are confusing me a bit more.

    I'm assuming this should be the correct integral \delta\pi25 \int_0^6{10-y}dy

    Which should be 65520 * pi
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    Re: Work and Fluid Force

    Quote Originally Posted by Hartlw View Post
    It's the work necessay to lift the water from it's cg to the top of tank: W=wh

    Or integrate over the work for each slice using the same formula.
    The first sentence is wrong. The second sentence is right, as already given by HallsofIvy.

    I belabor it because my wrong statement is interesting (a misconception):

    w= weight of water, W=total work to take water to rim, h=height of tank, H = height of water, ρ is weight density

    dw = ρAdx
    dW = (ρAdx)(h-x), work to take slice to rim.

    Integrating from 0 to H gives:

    W = wh – ˝ ρAH2

    hmmmm. this simple problem has led to a valuable lesson (for me):
    The work done in lifting a mass of weight w a distance h is wh only if the mass doesn’t change shape.

    Not entireley comfortable with that.
    Last edited by Hartlw; February 8th 2014 at 12:38 PM.
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